内外进行两次数论分块,可以做到求解
#include <cstdio>
#include <iostream>
using namespace std;
#define Mod 20101009
const int MAXN = 10000000;
int n , m , f[ MAXN + 5 ];
int k , prime[ MAXN + 5 ] , mu[ MAXN + 5 ];
bool vis[ MAXN + 5 ];
void sieve( ) {
mu[ 1 ] = 1;
for( int i = 2 ; i <= MAXN ; i ++ ) {
if( !vis[ i ] ) {
prime[ ++ k ] = i;
mu[ i ] = -1;
}
for( int j = 1 ; j <= k && 1ll * i * prime[ j ] <= MAXN ; j ++ ) {
vis[ i * prime[ j ] ] = 1;
if( i % prime[ j ] == 0 ) break;
mu[ i * prime[ j ] ] = -mu[ i ];
}
}
for( int i = 1 ; i <= MAXN ; i ++ )
f[ i ] = ( f[ i - 1 ] + 1ll * mu[ i ] % Mod * i % Mod * i % Mod ) % Mod;
}
int Quick_pow( int x , int po ) {
int Ans = 1;
while( po ) {
if( po % 2 )
Ans = 1ll * Ans * x % Mod;
x = 1ll * x * x % Mod;
po /= 2;
}
return Ans;
}
int inv( int x ) {
return Quick_pow( x , Mod - 2 );
}
int Get( int n1 , int m1 ) {
int d1 = min( n1 , m1 ) , Ans = 0;
for( int l = 1 , r ; l <= d1 ; l = r + 1 ) {
r = min( n1 / ( n1 / l ) , m1 / ( m1 / l ) );
Ans = ( Ans + 1ll * ( f[ r ] - f[ l - 1 ] + Mod ) % Mod * ( 1ll * ( n1 / l ) * ( n1 / l + 1 ) % Mod * ( m1 / l ) % Mod * ( m1 / l + 1 ) % Mod ) % Mod * inv( 4 ) % Mod ) % Mod;
}
return Ans;
}
int solve( int n , int m ) {
int Ans = 0 , d = min( n , m );
for( int l = 1 , r ; l <= d ; l = r + 1 ) {
r = min( n / ( n / l ) , m / ( m / l ) );
Ans = ( Ans + 1ll * Get( n / l , m / l ) * ( l + r ) * ( r - l + 1 ) / 2 ) % Mod;
}
return Ans;
}
int main( ) {
sieve( );
scanf("%d %d",&n,&m);
printf("%d\n",solve( n , m ));
return 0;
}